Next in the list (now that the motors are here) will be the PSU. I think I understand what I'm doing but I wanted to ask if someone could stop me before I make any (or at least too many) terrible mistakes :) I know this is a rehash of threads to probably occur once a fortnight but its for a reason. I'm setting out the details because it's not just about buying the right thing - I want to make sure my understanding of what I'm buying and why is correct.

I have these 3.1Nm steppers (http://cnc4you.co.uk/resources/60BYGH301B.PDF) and I plan to wire them as parallel: 4.2A, 3.2mH and 2.73V. Doing some magic calculations (based on magic spells found elsewhere on this forum) leads me to an operating voltage of 57.2V. The PSU would therefore need to provide 57.2V@12.6A. But wait...

I plan to buy PM752s (http://www.zappautomation.co.uk/pm752-microstepping-driver-p-409.html?cPath=9_3_4) which can provide, inter alia, 4.09A or 4.64A. In fact they support up to 5.2A but the motor has a maximum of 4.2A. This would mean the appropriate setting would be 4.09A therefore the PSU would now only need to provide 57.2V@12.3A. But wait, again...

Now as far as I understand it, I don't need to make the full 56.2V available, as nice as the idea might seem to me. There would be a performance drop but as the motors are probably outperforming the rest of the machine anyway, would this necessarily present a problem? Common figures seem to be 42V and 48V. Is dropping from 56V to 48V a travesty of epic proportions?

So what about the current? If needed, I could run two PSUs wired in parallel to provide more current (ie 2 x 42V@7A would effectively be able to provide 42V@14A). However, much as the full 56.2V may not be needed, so too the full 12.3A may not be needed. For an unregulated supply, the PM752 docs suggest the PSU might only need to supply 50%-70% of the motor rating. Picking an arbitrary 75%, the PSU would now only need to provide 42V@9.3A. While 50%-70% would actually be 6.2A-8.7A, I assume (no! not assumptions! nooooo!) that more safety margin is always better than less.

Zapp's SPS407 (http://www.zappautomation.co.uk/sps407-p-620.html?cPath=10) - with 42V@7A(continuous) and 9A(peak) - would work on its own using the 50%-70% figures but would need to be doubled up to make my fairly arbitrary 75% minimum - though if doubled then for the same price I could consider the SPS705, two of which would provide 68V@10A. I gather this could be magically fettled to provide the ideal 56V-57V at the slightly-less-than-ideal 10A, but would move the cost bracket from £60-£70 to £120-£140.

CNC4You's 48V@12A (http://www.cnc4you.co.uk/index.php?route=product/product&path=81&product_id=33), by contrast, would provide 97.5% and a few extra volts all on it's lonesome, but would still be under the ideal voltage.

I notice that Zapp also bundle their PS806 (http://www.zappautomation.co.uk/ps8065-linear-power-supply-p-518.html?cPath=10) 68V@6A with 3xPM752s and 3Nm motors, but that 6A falls below the 50% so I am presuming it would be too far a stretch in this very-similar-but-a-tiny-bit-different scenario.

Then we move onto sourcing the bits and building one but for some reason that makes me a little jittery compared to getting something made by someone who is competent to build something I plan to plug into the mains!

All of this leads to a point, eventually. If I choose "money" then it's 48V@12A@£57. If I choose "voltage" then it's 56V@10A@£138.

Am I on the right tracks and am I considering the right things to balance? I've picked up this idea that I can run on less volts, but am I being daft (well, the anwer to that is always yes) to think I should be considering anything less than 56V in this scenario?

Jonathan

11-09-2012, 05:48 PM

Looks like you're after an explanation of why, not what...

I would consider the voltage you get from the common formulas (20*V, 32*L^0.5 etc) to be a 'recommended minimum'. You will get better torque at high speed if you use 70-75V, plus seems a waste to get PM752 drivers if you're only going to run them on 42V.

A linear power supply handles current surges better than a switch mode power supply (smps), which means you can safely use a lower rated current than the smps. A 500VA transformer is fine for 3 motors, probably 4 and that's only 500/70=7.1A.

The current in a 2-phase stepper motor on a microstepping driver is, at intermediate speed, approximately sinosoidal and out of phase by 90°. This means that the maximum current is the maximum of I*(sin(thing)+cos(thing)), which is I*2^0.5 where I is the current in one phase - so 2.1*1.414=2.97A. This is 70.7% of 4.2A. Similarly if you work out the rms current for both phases (just integrate the previous formula squared etc), it is 2.1A - hence 50%. I'm guessing this is where your 50%-70% approximation comes from.

JAZZCNC

11-09-2012, 05:52 PM

Ok the calcs you've done are close and correct in some ways but some of what you read I think you miss understood.? The 752s Manuel and 50-70% was referring to current not voltage. So you still need 57V but only 8-9A current.

That said has you know the voltage is what gives you the speed in a stepper so dropping to 48V means your only getting 85% of the speed you would at 57V. PLUS these motors will happly run at 70V thou I always run them at 65-68V to give a better margin for Back emf. So really at 48V your only getting 69% of the motors potential at 70V.?

Now in speed terms to the machine it can be dramatic in terms of the motors corner speed or point torque starts dropping away.
IE @48v then expect the corner speed less than 700rpm and @70v approx 1000rpm.
On a 5mm pitch screw then it makes the usable feed rate difference between @48=3500mm/min and @70v=5000mm/min.

Without knowing more about pitch and what you want to cut then hard to say best route to take.? .. BUT . If the voltage can provide the speed in combination with your screws to give the feed rates you need to cut what you want then it's not an issue really simple has that.!!

Remember high rapids mean pretty much nothing in real world cutting unless you intend to do lots of positional moves like drilling or have a massive long machine.? . . Mostly just bragging rights.!!

That said I prefer building my own supplies so obviously can build it to my exact needs so nearly always 65-68V using 75V drives simply because I can.! The amps just depends on how motors and number of them.
It's easy building a toroidal PSU and often cheaper plus safer for the drives has they handle back Emf much better and gives a nice consistent supply of power.

For me it would be "Build own to my exact needs" other wise if speeds/feeds it allows are OK use the 48V@12A and save the money.! Certainly wouldn't buy 2x68V@5a has could easily build for less than price one.!!!

Rogue

11-09-2012, 06:50 PM

Looks like you're after an explanation of why, not what...

I would consider the voltage you get from the common formulas (20*V, 32*L^0.5 etc) to be a 'recommended minimum'. You will get better torque at high speed if you use 70-75V, plus seems a waste to get PM752 drivers if you're only going to run them on 42V.

I think that's it. It's easy to ask for the answer (and I often do) but it's sometimes better to ask for correction as a learning exercise.

I agree about the drivers. It was my understanding that the system could be run with lower voltage, I was trying to establish if doing this would lead to an earthshattering change or just knock a few percentage off the numbers. I would prefer to go for the higher-spec drivers and utilise them as best I can.

Ok the calcs you've done are close and correct in some ways but some of what you read I think you miss understood.? The 752s Manuel and 50-70% was referring to current not voltage. So you still need 57V but only 8-9A current.

Yep I got that bit, though I guess I didn't really make it clear. The reduction to 42V or 48V wasn't because I was applying a percentage reduction, I was just observing that several of the CNC places were offering that size of PSU and if I wanted to buy off the shelf then I was limiting myself to what was on the shelf in the first place!

Without knowing more about pitch and what you want to cut then hard to say best route to take.? .. BUT . If the voltage can provide the speed in combination with your screws to give the feed rates you need to cut what you want then it's not an issue really simple has that.!!

Ahah. Of course, the benchmark against which the performance needs to be measured. I was asking the wrong questions again! Not "will it work" (because the answer can only be "it depends") but "what do I need it to achieve?"

For me it would be "Build own to my exact needs" other wise if speeds/feeds it allows are OK use the 48V@12A and save the money.! Certainly wouldn't buy 2x68V@5a has could easily build for less than price one.!!!

I keep looking at it and thinking "that looks quite straightforward", but then I think about plugging it into the mains and I have a vision of mushroom clouds rising over Derby. Some might say it would be an improvement, of course.

I'm going to end up trying to build this, aren't I? Bugger. Well, most of the components are cheap enough,

JAZZCNC

11-09-2012, 06:58 PM

I'm going to end up trying to build this, aren't I? Bugger. Well, most of the components are cheap enough,

Yep probably.?? BUT it is Honestly so easy you'll wonder why you fretted afterwards and if you need a hand holding then just shout.!

Rogue

11-09-2012, 07:34 PM

So...

...this 500VA 35V-35V transformer (http://www.rapidonline.com/Electrical-Power/Toroidal-Transformer-500va-0-35v-0-35v-88-3839) would let me connect the two outputs in series to get 70V@7.1A. By the time it has lost a few volts through the additional components (rectifier etc) I would have 68.6V (I think it's 1.6V drop for suitable rectifiers?). This 30V-30V (http://www.rapidonline.com/Electrical-Power/Toroidal-Transformer-500va-0-30v-0-30v-88-3838) would give me 58.4V - with 57.2V being the minimum rather than the optimal that I had thought it represented - so it would be the bottom line option.

Daft question, is it feasible to knock the output down a bit from a transformer, short of fiddling with the number of turns?

JAZZCNC

11-09-2012, 07:44 PM

No the 35V is AC so needs to be rectified which makes it times 1.4 so that would be 35 +35=70 x 1.4= 98Vdc. . .Boom.!!

You need 25V to give 70Vdc.

Yes can fiddle with turns but why would you.? Just buy the appropriate transformer.??

Rogue

11-09-2012, 07:47 PM

Yep probably.?? BUT it is Honestly so easy you'll wonder why you fretted afterwards and if you need a hand holding then just shout.!

I fret because I'm used to being an expert in my chosen field and here I'm dealing with things that I have absolutely no experience (or confidence) in. I am very good at imagining all the horrible ways things could go tragically wrong, though. Each action taken or order placed commits scarce resources, and every mistake is a painful waste. Otherwise I'd quite happily glue everything to everything else and pump lightning through it to see what would happen :whistle:

Hand holding? I would rather throw my ideas out and be corrected so that my understanding improves, but I'm sure some handholding creeps in unwittingly. My apologies if it does because that's not what I want to do.

Rogue

11-09-2012, 07:51 PM

No the 35V is AC so needs to be rectified which makes it times 1.4 so that would be 35 +35=70 x 1.4= 98Vdc. . .Boom.!!

You need 25V to give 70Vdc.

Yes can fiddle with turns but why would you.? Just buy the appropriate transformer.??

Ahh, yeah that might not be what I'd want then. I was asking about the fiddling because I was wondering, I certainly don't want to fiddle...

Jonathan

11-09-2012, 08:00 PM

Ahh, yeah that might not be what I'd want then. I was asking about the fiddling because I was wondering, I certainly don't want to fiddle...

You can wind a few extra turns on, or vice versa, then add the appropriate insulation. It's a crude but successful way of tweaking the voltage of a toroidal transformer that's not quite what you want...it shouldn't be needed here though as you can just buy the right one.

JAZZCNC

11-09-2012, 08:07 PM

Hand holding? I would rather throw my ideas out and be corrected so that my understanding improves, but I'm sure some handholding creeps in unwittingly. My apologies if it does because that's not what I want to do.

Yep I understand but Wasn't offering to do it for you just over look and backup.!! An experienced eye watching over never hurts only affirms you've learnt correct.

Rogue

11-09-2012, 08:08 PM

Ahh hold on, wait a second. It would be 25V if the outputs were connected in series (thereby doubling the V while maintaining the A), so 25+25*1.4=70V@8.9A. If a transformer with 50V-50V 625A (http://www.rapidonline.com/Electrical-Power/Toroidal-Transformer-625va-0-50v-0-50v-88-3844) had the outputs wired in parallel, would this give 50V*1.4=70 @ 625+625 / 70= 17.8A?

JAZZCNC

11-09-2012, 08:15 PM

Yes 50v wired parallel is double the amps but they are half the rating of the 25V so workout the same.

Rogue

11-09-2012, 09:00 PM

Could you just clarify what you mean by rating in that context? VA? I ask because the only way I understood that the 25V wired in series would equal the amps from the 50v wired in parallel would be for the 25V to start with double the VA, ie

I really appreciate both of your inputs, time and patience, just in case I didn't mention it already!

Jonathan

11-09-2012, 09:17 PM

When you put the windings in series add the secondary voltages, but at the same current. In parallel you get the same voltage as for one winding but twice the current. Either way the VA rating has to remain constant for the same transformer.

JAZZCNC

11-09-2012, 09:29 PM

Just what Jonathan says but to make it clearer look at the transformer output current ratings for these two transformers.

If you wired the 2x25V in parallel not series you'd still have 25V but @ 24A So obviously 2x50V in paralle is 50V @12A (6A x 2)

Rogue

11-09-2012, 09:37 PM

Now, are you agreeing with me or correcting me? Because what you said is what I thought I said, if you follow me.

Rogue

11-09-2012, 09:45 PM

Ahh, wait a second. Is the 625VA shared between the two outputs or per output?

JAZZCNC

11-09-2012, 09:47 PM

Now, are you agreeing with me or correcting me? Because what you said is what I thought I said, if you follow me.

Who you asking Jon or Me.? Either way we are both saying series you 2 x volts 1xamps or parallel 1 x volts 2 x Amps. . . BUT . . the amp output rating for 25v transformer are double the 50V.!

Rogue

11-09-2012, 10:09 PM

So we were saying mostly the same thing. That's a good start :)

I think I've worked out where my mental block is for this element at least. I was treating parallel as increasing A rather than decreasing the division of A. In other words, I was thinking of two outputs, each with 50V 625VA. Parallel would make this one output of 50V 1250VA. Rather, I should be looking at it as two outputs of 50V sharing 625VA between them - effectively making it 100V / 625VA = 6.2A. Wiring it in parallel would then have the same VA but divided by less, which has the same end effect as doubling it: 50V / 625VA = 12.5A. When I look at it like this, your previous comments make more sense so I guess I'm on the right track.

And you wonder why I get jittery with mains power :joker:

JAZZCNC

11-09-2012, 10:12 PM

Ahh, wait a second. Is the 625VA shared between the two outputs or per output?

Ok I see where your going wrong.? Nope forget the Va rating for now the only bit your interested in is the secondary outputs.

You have 2 and you can either have them separate so effectively making 2 completely separate PSU's or wire them together to make one large supply.
If wired in series then you add the voltages for each secondary together but keep the amps the same. Wire in Parallel then you do the opposite.

This bridge has a forward voltage drop of 1.1v per diode so your output volts will be 50 X 1.4 - 2 x 1.1 = 68v! Also each diode will dissipate 1.1v * 12A = 13.2W so you'll have 26W of heat to dissipate. Assuming air temp of 25degC, and a rectifier temp of 125degC you need a thermal resistance of no more than (125-25)/26 = 3.8degC/W, since the rectifier has a thermal resistance of 1.4degC/W from the diode to the mounting base it'll need to be mounted on a heatsink rated at 2.4degC/W or better. Something like this: Buy Heat Sinks Heatsink 2.4K/W 97x50x25mm ABL Components 333AB0500B online from RS for next day delivery. (http://uk.rs-online.com/web/p/heat-sinks/0271836/) @ £3.79 and don't forget the thermal compound between rectifier and heatsink and between heatsink and the case...

Doing this in one capacitor is expensive (18000uF, 400v, 12A ripple = £271!!!) So usually you make it up with a bank of capacitors in parallel. The exact value is not critical so 3 x 5000uF @ 100v/4A would do, or whatever you can get cheap on eBay :)

For example, 3 of these: Buy Aluminium Capacitors GP Al electrolytic capacitor,4700uF 100V Epcos B41456B9478M online from RS for next day delivery. (http://uk.rs-online.com/web/p/aluminium-capacitors/2550282/) would be £33, with the matching mounting clips (http://uk.rs-online.com/web/p/capacitor-mounting-clips/2035698/) @£1.70 for a pack of 5. These have screw terminals so you'll need the appropriate solder ring tags.

You'll also need a bleed resistor across each capacitor to discharge it when the power is turned off (for safety). The energy stored is 0.5CV^2 joules. For each capacitor spec'd above, the energy is 0.5 * .005 * 70^2 = 12joules. A watt is a joule per second, so for a 5 second discharge you need to dissipate 12/5 = 2.4W so we're going to need 3 or 5W rated resistors. Assuming the resistor dissipates the same 2.4W when the power supply is turned on the value will be R = v^2/W = 70^2/2.4 = 2000ohm. I'd use 2200ohm (2k2) @ 3W wire ended and solder them directly to the ring tags on the capacitor: Buy Through Hole Fixed Resistors ROX3S metal oxide film resistor,2K2 3W TE Connectivity ROX3SJ2K2 online from RS for next day delivery. (http://uk.rs-online.com/web/p/through-hole-fixed-resistors/2142803/). (http://uk.rs-online.com/web/p/through-hole-fixed-resistors/7078892/) (pack of 10 @ 99p)

Doing this in one capacitor is expensive (18000uF, 400v, 12A ripple = £271!!!) So usually you make it up with a bank of capacitors in parallel. The exact value is not critical so 3 x 5000uF @ 100v/4A would do, or whatever you can get cheap on eBay :)

For example, 3 of these: Buy Aluminium Capacitors GP Al electrolytic capacitor,4700uF 100V Epcos B41456B9478M online from RS for next day delivery. (http://uk.rs-online.com/web/p/aluminium-capacitors/2550282/) would be £33, with the matching mounting clips (http://uk.rs-online.com/web/p/capacitor-mounting-clips/2035698/) @£1.70 for a pack of 5. hese have screw terminals so you'll need the appropriate solder ring tags.

Thank you, that's very, very useful. I was looking at caps that cost about £3 each (http://www.rapidonline.com/Electronic-Components/4700uf-100v-Snap-in-Electro-Capacitor-11-2912) though, is there an important part of the spec that I didn't take into account?

Jonathan

12-09-2012, 10:54 AM

This bridge has a forward voltage drop of 1.1v [...] each diode will dissipate 1.1v * 12A = 13.2W so you'll have 26W of heat to dissipate.

The mean current will be significantly less than 12A, so a heatsink with a much higher thermal resistance could be used. On my stepper PSU, which also uses a 500VA 50V transformer with 3 motors, I used one of these (http://www.rapidonline.com/Electronic-Components/8A-SIL-Bridge-rectifiers-78795) (or this one (http://www.rapidonline.com/Electronic-Components/35A-Bridge-rectifiers-29671)) and no heatsink is required as I've never noticed the rectifier get 'very warm' let alone 150°C. Still, you can find good heatsinks for free in lots of things, or since this evidently doesn't need much you could just attach it to a reasonable size sheet of aluminium.

Agree with the calculation for the bleed resistor, but surely since the stepper drivers will be switched on until the voltage has dropped to about 20V, resistors aren't critical since the energy from the capacitors will dissipate into the motors? If you unplugged the stepper drivers before turning off the power supply then the resistor is needed, but that would be a strange thing to do since it risks breaking the drivers.

Thank you, that's very, very useful. I was looking at caps that cost about £3 each (http://www.rapidonline.com/Electronic-Components/4700uf-100v-Snap-in-Electro-Capacitor-11-2912) though, is there an important part of the spec that I didn't take into account?

The ones you have linked to will be fine, I'd get 4. The ones irving linked to are better quality (lifetime etc), but cheap capacitors are fine for this application.

Rogue

12-09-2012, 01:24 PM

The ones irving linked to are better quality (lifetime etc), but cheap capacitors are fine for this application.

Yeah, he likes to give me expensive advice. I'm still trying to work out how to safely situate this machine after he got me to check the rafters :whistle: Keep an eye out for the upcoming threads on building waterproof, soundproof and giant-maneating-spider-proof low profile enclosures with built in dehumidifiers....

Not that I mind in the least of course. I'd rather have the advice to weigh up than not have it and blunder on blindly.

I'm intrigued though. What kind of thing would you be doing to benefit from the price difference?

JAZZCNC

12-09-2012, 05:14 PM

I'm intrigued though. What kind of thing would you be doing to benefit from the price difference?

Well if this machine was working 14hrs days 6 days a week cutting £500 lumps of Ali then you'd want to know it's reliable and not going to break down just for the sake of £30.!!

Edit: Jons right thou for your use then they are fine and actually I have the same ones on my machine.!

JAZZCNC

12-09-2012, 05:21 PM

Yeah, he likes to give me expensive advice. I'm still trying to work out how to safely situate this machine after he got me to check the rafters :whistle: Keep an eye out for the upcoming threads on building waterproof, soundproof and giant-maneating-spider-proof low profile enclosures with built in dehumidifiers....

You forgot " That are electricly safe". . :)

Rogue

12-09-2012, 06:44 PM

You forgot " That are electricly safe". . :)

Actually I'm counting on the sparks, arcing and random live surfaces to keep the giant-maneating-spiders away if I end up having to use the basement...

irving2008

12-09-2012, 08:44 PM

Yeah, he likes to give me expensive advice. I'm still trying to work out how to safely situate this machine after he got me to check the rafters :whistle: Keep an eye out for the upcoming threads on building waterproof, soundproof and giant-maneating-spider-proof low profile enclosures with built in dehumidifiers....

Not that I mind in the least of course. I'd rather have the advice to weigh up than not have it and blunder on blindly.

I'm intrigued though. What kind of thing would you be doing to benefit from the price difference?

The cheap capacitors are a bit marginal on ripple current for three but as Jazz/Jonathan says they'll do the job for the level of use you'll put them to esp if you use four of them. I did qualify my statement with the phrase... "or whatever you can get cheap on eBay :)" and they were examples. I have 2 x 7500uF 400V capacitors on my 68v supply, I paid £5 each on eBay - they list at £75 each, used in hi-quality 500W+ sound systems - the D-class output stage of a high-power switching amp looks remarkably similar to a stepper drive :). The only thing I don't like about the ones you've chosen are the solder terminations; its a personal preference but I like meaty screw terminals.

On other points...

Bleed resistors are there for safety. You might never power the supply on without the steppers connected but I can't know that and I would never advise building a hi-voltage (>48v) supply without them. Leaving them out isn't best practice (it ranks with wiring e-Stops with 230v and having plugs on the 'live' side in my book) and I've been caught out by that before. For the record, I've been in electronics manufacturing on and off for over 30y and have seen most levels of stupidity :)

As regards the heatsink Jonathan, of course the case doesnt get to 150degC, its what the junction mustn't go above (actually 125degC for your device). If you look at the datasheet for the rectifier you used its rated at 35A @ 55degC case temp. At 35A its dissipating 35 x 1.1 x 2 = 77W so given the limiting junction temp is 125degC the internal thermal resistance must be (125-55)/77 = 0.9degC/W therefore case to ambient of 25degC must be 0.4degC/W to run with no heatsink (which is what the datasheet seems to suggest, though I find that very low, but it is a metal case so go figure). At 12A (which I accept is a worst case with all three motors at full chat, relatively rare) your case temp would be 35degC. On the face of it thats a better rectifier for this purpose. The one I suggested has an epoxy case so clearly needs the heatsink as detailed in its datasheet.

JAZZCNC

12-09-2012, 09:07 PM

Bleed resistors are there for safety. You might never power the supply on without the steppers connected but I can't know that and I would never advise building a hi-voltage (>48v) supply without them. Leaving them out isn't best practice (it ranks with wiring e-Stops with 230v and having plugs on the 'live' side in my book) and I've been caught out by that before. For the record, I've been in electronics manufacturing on and off for over 30y and have seen most levels of stupidity :)

Couldn't agree more and here's an example of why.!
I've just acquired an old cnc lathe out of a school and been ripping it's guts out has it runs a bespoke control/software and I want to run Mach3.
This thing has 3 boards Control boards, PSU board and spindle board. I'm keeping the PSU and Spindle so after pulling the control board out I was messing with the spindle board trying to work out what wire goes where when I got a pretty wicked kick I wasn't expecting seen has it wasn't plugged into mains.?? . . . .Yep it's got toroidal PSU with no Cap drain.

Not enough to hurt me but woke me up good and that was only small PSU running at 36V ish with relatively small caps.!!

SO I completely agree with Irving and always build them into my PSU's..!! . . Now the only thing I don't like about Irvings way is that the resistors are permanently draining while the machines running, this creates heat while running so warms the control box and therefore the drives etc.
I'm a bit Anal with my control box and build them with relays and full safety using 24V and control lots of things with relays, one of these is a NC relay which kicks in and out a resistor so it only drains when off.!! . . No such thing has being too safe IMO.

Rogue

12-09-2012, 09:10 PM

For the record, I've been in electronics manufacturing on and off for over 30y and have seen most levels of stupidity :)

And I'm grateful for the chance to be able to draw on that experience, along with the experience and knowledge that Jazz and Jonathan have brought to the discussion.

Fuses - now that never occured to me but sounds like a rather good idea...

irving2008

12-09-2012, 09:48 PM

...

Fuses - now that never occured to me but sounds like a rather good idea...
Err.. yes. Fuse the primary side with a slow-blo fuse rated approx 50% - 100% more than the likely operating current... for a 625VA transformer thats a 4 or 5 amp fuse. You can fuse the secondary with a 15A slo-blo as well if you want. Put the fuse between the transformer and the rectifier.

Jonathan

12-09-2012, 10:17 PM

You might never power the supply on without the steppers connected

To be fair it's a good idea to power the supply on without the steppers connected when you've first built it to check the voltage is correct, so yes have the bleed resistor.

As regards the heatsink Jonathan, of course the case doesnt get to 150degC, its what the junction mustn't go above (actually 125degC for your device).

I should have clarified I've used both on different power supplies. It's 150°C for the first one I linked to and 125°C for the other.

I just put an ammeter on my 500VA PSU and ran a few programs on the router. With 4 motors maximum current was 2.66A and at standstill 0.72A. Just going by eye (not great I know, should use my oscilloscope) the mean current looked to be about 1.9A. The drivers are set to 50% current at standstill, so since the power supply is outputting 68V we expect the current current when stationary for 4 motors to be 4*0.5*2.73*4.2/68*2=0.674A, so the measurement is fairly close.

So lets be very safe and calculate it with the peak current since we know if the temperature is fine at that current (2.66A) it must be fine for the lower current when running. The 8A rectifier I used is 2.2°C/W junction to case and forward voltage is 1V, so 2.66*1=2.66W to dissipate. (150-25)/2.2=68°C/W which is a large value, hence why it works with no heatsink. Also the datasheet specifies up to 3.2A with no heatsink, so clearly we're fine especially as in reality the mean current for 3 motors is less than 2.66A.

Edit: Just tried measuring using the oscilloscope with a shunt resistor and running one motor at the speed it draws the highest current I get 1.58A mean, so 4.75A for 3 motors which implies if I get all 4 motors running at their peak, it will draw 430W from the 500VA transformer.

JAZZCNC

12-09-2012, 10:38 PM

Don't all these figures make your head hurt.?? . . . . For sake of £1 Just stick it on a heat sink to be sure then it's done and forgot about.!!

Jonathan

12-09-2012, 10:44 PM

Don't all these figures make your head hurt.??

No, they make life more interesting.

Here's a graph to brighten things up:

6849

(0.47 ohm shunt resistor and one motor going at speed which obtains the highest current)

JAZZCNC

12-09-2012, 11:04 PM

No, they make life more interesting.

Love pretty pictures but told you before you need to get out more pal. .:couple_inlove:

Rogue

12-09-2012, 11:07 PM

SO I completely agree with Irving and always build them into my PSU's..!! . . Now the only thing I don't like about Irvings way is that the resistors are permanently draining while the machines running, this creates heat while running so warms the control box and therefore the drives etc.
I'm a bit Anal with my control box and build them with relays and full safety using 24V and control lots of things with relays, one of these is a NC relay which kicks in and out a resistor so it only drains when off.!! . . No such thing has being too safe IMO.

I've read elsewhere about people using lamps rather than resistors as a "visual" indicator, is this the same kind of thing? I'm intrigued if a lamp connected to a normally closed relay would work, though I have no intention of trying it out...

Jonathan

12-09-2012, 11:16 PM

I've read elsewhere about people using lamps rather than resistors as a "visual" indicator, is this the same kind of thing? I'm intrigued if a lamp connected to a normally closed relay would work, though I have no intention of trying it out...

Don't see why not, but bear in mind a lamp can blow/fail - so even if it's not illuminated you don't know for sure that there's no voltage across it.

JAZZCNC

12-09-2012, 11:20 PM

Don't see why not, but bear in mind a lamp can blow/fail - so even if it's not illuminate you don't know for sure that there's no voltage across it.

Took the words right out my finger tips.! . . Resistor is a safer bet, could always use a resistor and a bulb but why complicate things.

irving2008

13-09-2012, 07:08 AM

Again, its good practice to have 'voltage present' indicators, especially if you have a secondary fuse. (plus I like lots of lights :) ).

For the primary side a panel mounted neon indicator w (http://www.rapidonline.com/Electronic-Components/Push-fit-12mm-Neon-Indicators-518519)ired directly across the transformer primary.

For the secondary side a 5mm LED (http://www.rapidonline.com/Electronic-Components/Kingbright-LED-Standard-5mm-77692)mounted in a panel mount (http://www.rapidonline.com/Electronic-Components/Lens-mounts-for-3-5mm-LEDs-61711) with a 3k9 2W resistor wired across the output.

Rogue

13-09-2012, 02:28 PM

You'll also need a bleed resistor across each capacitor to discharge it when the power is turned off (for safety). The energy stored is 0.5CV^2 joules. For each capacitor spec'd above, the energy is 0.5 * .005 * 70^2 = 6joules. A watt is a joule per second, so for a 5 second discharge you need to dissipate 6/5 = 1.2W so we're going to need 2W rated resistors. Assuming the resistor dissipates the same 1.2W when the power supply is turned on the value will be R = v^2/W = 70^2/1.2 = 4000ohm. I'd use 3900ohm (3k9) @ 2W wire ended and solder them directly to the ring tags on the capacitor: Buy Through Hole Fixed Resistors Carbon Resistor, 2W ,5%, 3k9 RS RS-Carbon-3k9-5%-2W online from RS for next day delivery. (http://uk.rs-online.com/web/p/through-hole-fixed-resistors/7078892/) (pack of 10 @ 72p)

After a cup of tea and a bit of a read, I have a quick question in relation to 0.5CV^2. Where you give 0.5 * 0.005 * 70^2 = 6, I end up with 12.2. In fact using the time honoured technique of "randomly changing numbers until they worked", I only got 6j by changing the 70v to 16v?

I'm rather hoping you made a typo because the alternative is that I'm more stupid than I thought!

irving2008

13-09-2012, 09:21 PM

After a cup of tea and a bit of a read, I have a quick question in relation to 0.5CV^2. Where you give 0.5 * 0.005 * 70^2 = 6, I end up with 12.2. In fact using the time honoured technique of "randomly changing numbers until they worked", I only got 6j by changing the 70v to 16v?

I'm rather hoping you made a typo because the alternative is that I'm more stupid than I thought!

Errrmmm... looks like I made a boo boo... :hororr: might have multiplied by 0.5 twice in error... your calculation is correct... now I'll go and edit my post..

BTW 2k2 @ 3w will be fine

Rogue

13-09-2012, 10:03 PM

Then I shall take off my dunce's cap and come out of the corner. For now, at least :whistle:

m.marino

15-09-2012, 11:04 AM

Okay,

I have been reading and talking to folks (Thanks Jazz) on the power supply issue. This is what I understand and what I want to make sure of.

Mains power (fused at plug) To
Fused Switch (main Switch as will wire in a secondary for the motors) This is feed to a + rail - rail and GND
Positive rail Feeds:

A Circuit Breaker going to the Switch for the Soft start for the transformer (will be asking on how to build one of those)
The PMDX-126 BOB (Which has built in Circuit Breaker with auto reset)
The on board relay (on the PMDX-126) for the spindle (30A relay) Currently this feeds a Kress 1050 FME. Does this need a Circuit Breaker as well?

Negative Rail Feeds same as listed above.
GND includes all of above as well as grounding the shielding on the stepper cables, the computer frames (control box and computer), Machine frame, and the unshielded side of the PMDX board.

Okay that is so far, now onto the power supply that is going to be going in shortly. The way I am looking at setting it up is as follows:

From Circuit breaker in number 3.1 to Switch for motor power (both +/- plus GND if that seems best)
From there Negative and GND go to the Soft start for the transformer (need help on designing one of these so that I can build it in)
Positive goes to a Relay (12V) attached to the PMDX in the set up in NO position so that the motors cannot be powered unless the board has power.
From Soft start to 750VA 50+50 Standard Range Toroidal Transformers: CM0750250: 750VA 230v to 2x50v (http://www.airlinktransformers.com/chassis_mounting_toroidal_transformers_standard_ra nge/3-CM0750250.html) This is to power four 4.2A peak current Nema23 steppers and by what I have read, should meet even heavy running requirements.

I am going with the 7.5A 70V in Parallel as that gives my 15A output in theory minus any loss and should more then cover the 12.3A needs to drive the four stepper motors (I recall from Physics that current is on demand and Voltage you get what is there, a 3A item only draws 3A but will take all the Voltage that comes with those 3A).

This is where I am a bit confused. should I put another fuse here between the Transformer and the Rectifier or should I run directly to them?
The Rectifier is a 35A 200V Metal casing rectifier and will be mount to a ali' plate to help with heat transfer and I have thermal paste (does the bolt/post need to be isolated?).
Next is The Capacitors and I am looking at 5 4700 uF units (23.5 uF to meet a need for 21.4uF)
Okay I am looking at this and trying to figure out how to wire in the the resistor to drain off power when the system shuts off. The idea I have is to put a relay in place that is in the NC position with the resistor (50W as they are not expensive and give some added safety margin) set to drain off the power when the switch closes. Now I could power this relay from the PMDX as with the other relay in place the motors should not normally be able to be powered if the board power is off. Other then that I am at a dead end on this one. Also the resistor should go to ground when it is activated correct? Also could use a double check on the Ohms I need for that Wattage.
After the Capacitors the power is then run to a common rail from which the fuses (5A) for each Driver is in line with the + and the _ runs back to a common.

If that all makes sense and anyone wishes to help with the Soft start modules and other questions I have posted here in, please do. Also hopefully this will help some other folks with design and structure building. I plan on putting this all on a board and Then mounting on standoffs in the control box. Yes it is bit of over kill but then again I make a part of my living from this machine and it needs to be safe and run well.

For those who want to get a bigger picture of the electronics of this System I am using a AMD based PC running XP Pro SP3, using Mach3, That goes to a Smooth Steeper Ethernet card and then to the PMDX-126. Which then passes the data onto AM882's Digital drivers. This drives 2 motors on X, and one each on Y and Z axis. Spindle is currently a Kress 1050 FME. I have Jazz, James, and a few others to thank for aid in the development of this machine and in the new thread it passed the test you can see a picture of the detail it can do. The reason for the new power supply is that I owe the one I am using back to the person who has so kindly let me use it this long.

Michael

edited to correct a typo

m_c

15-09-2012, 01:08 PM

Soft start for that size of transformer isn't really needed. Just use a zero volt switching SSR (most AC SSR's are), and you shouldn't have any issues.
One thing to note though, is if the SSR fails, it'll likely fail short circuit, so oversize and fuse it to minimise the risk. If you're really concerned, put a mechanical relay in circuit before it, incase anything does happen.

Generally you only fuse the transformer primary side. Any overload on the secondary side should blow the fuse on the primary side. If you oversize the rectifier enough, then any overload on the DC side should also blow the primary side fuse.
Fusing individual drives isn't usually recommended, as if it fails for any reason, then you risk the drive being fried due it having no where to dump power from a moving motor to.

The drain resistor should simply connect across the capacitor terminals. Unless you have some reason to quickly dump power, you don't need to have it switched. The motor drivers themselves will quickly drain the capacitors with power removed.

irving2008

15-09-2012, 02:18 PM

Mike,

Can I suggest you get some terminology correct, it might save you some pain (physically and metaphorically) later

When referring to AC (mains) its Live, Neutral, Ground The colours being Blue, Brown, Yellow/Green

When referring to AC (transformer secondary) its AC Secondary and general practice to use Brown

When referring to DC then its +/positive, -/negative, chassis ground and colours are generally Red, Black, Green

Here's a sample circuit with estop and interlocks. Turning on the main switch powers up the +24 volt safety circuit. Push to start closes relay 1 (AC side) and powers up the +70v and aux supplies. Relay 2 (LV side) also closes holding in the relays unless the eStop is triggered. As an example I've shown one set of Realy 2 contacts switching the bleed resisors though I personally don't bother and have them permanently connected.. Other arrangements are possible...

6881

m.marino

15-09-2012, 02:38 PM

Irving my apologies for not using the correct terms. I know the AC usage and colors (Though I still prefer the older colors as they made much more sense to me).

Michael

Rogue

15-09-2012, 07:34 PM

Just out of interest, if you wanted to get 5V out of that system as well, would you run a 7805 off the +24v circuit, or should that circuit be kept purely for the relays/e-stop?

irving2008

15-09-2012, 10:09 PM

Just out of interest, if you wanted to get 5V out of that system as well, would you run a 7805 off the +24v circuit, or should that circuit be kept purely for the relays/e-stop?

Depends. The 24v is present as soon as the primary switch is turned on. If its OK for the 5v to be present too, and the current drawn is small, say < 0.1A, then yes. Otherwise I'd run a small 5v supply off the aux AC.

Intriguing, and hopefully stripboard friendly! I think I will start here before attacking the bigger stuff.

Here we are talking about mA: would the same 35A bridge resistor still work? Or would it be too "robust" (for lack of knowing what the appropriate term may be) in this situation? Does there come a stage when the safety margin becomes so excessive that it degrades performance at the lower end of the operating range?

irving2008

15-09-2012, 11:43 PM

Intriguing, and hopefully stripboard friendly! I think I will start here before attacking the bigger stuff.

Here we are talking about mA: would the same 35A bridge resistor still work? Or would it be too "robust" (for lack of knowing what the appropriate term may be) in this situation? Does there come a stage when the safety margin becomes so excessive that it degrades performance at the lower end of the operating range?

You mean 35A bridge rectifier. It would but one that size is needless. This one (http://www.rapidonline.com/Electronic-Components/Woo5-1-5a-50v-Bridge-Rectifier-47-3190) is more than sufficient and cheaper.

Rogue

19-09-2012, 10:49 PM

Here's the plan - listing is indicative of order, should roughly correlate to this:

6926

It's a little rough, I rushed it for the post. You might need to pretend there is a relay-controlled bleeder. And you might need to pretend there are any outputs to the rest of the world, but hey, that's the power of i-m-a-g-i-n-a-t-i-o-n!

Transformer for 70V system - Rapid has 500VA 2x50V transformers (http://www.rapidonline.com/Electrical-Power/Toroidal-Transformer-500va-0-50v-0-50v-88-3840) which should give me 50VDC@10A in parallel. I'd prefer 625VA but they don't carry 2x25 and the 2x50 has a 1-2 month wait from order. As I'll be sourcing this component last I have time to hunt around. A few other links to sources were posted in the thread; I'll check up on them later.

The wires to the relay that turns on the 70V will pass through another, normally open relay. This second relay is powered

Various crimps, wire and connectors should be easily sourceable at the local shop.

If I don't post for a month, my widow will probably be selling my stuff on ebay so keep an eye out for some bargains - she might be selling it for what I told her it cost before I got turned into crispy bacon...

irving2008

19-09-2012, 11:26 PM

For safety reasons don't use a toggle switch for the 'start' switch on the 24v supply to relay 1. It must be a momentary push to make to enable both relays and use a contact on relay 2 to lock it in, as I drew it.

Why? your machine hits a limit switch, triggering estop (or you hit estop). everything turns off. You turn off the power at the main isolator but in your desire to see what happened you forget to turn off the 'start' switch. You reset the estop, or move the machine off the limit switch and turn on the main isolator... but the start switch is still closed so the 70v rail comes up and the machine starts moving again because you forgot to stop the PC...

Its a safety interlock, it mustn't remain in an ON position. Also you don't need both contact on Relay 2 to drive Relay 1, only one side of the coil needs to be switched.

JAZZCNC

19-09-2012, 11:30 PM

Using my imagination I've imagined some of that same 24v running thru E-stop going thru limit switch's as well and the RL2 relay having another contact for the 5v signal from BOB.!!

Edit: Oh and you'll need an limit override momentry button in there as well to allow reversing off the switch's.

Rogue

20-09-2012, 08:23 AM

For safety reasons don't use a toggle switch for the 'start' switch on the 24v supply to relay 1. It must be a momentary push to make to enable both relays and use a contact on relay 2 to lock it in, as I drew it.

Ahhh. The reason why I basically re-drew your drawing was to get my head around what was happening. Now you put it like that I can see what is happening there. I guess I should have asked. It's not like it's the first daft question I've had :)

Also you don't need both contact on Relay 2 to drive Relay 1, only one side of the coil needs to be switched.

I know, but I wasn't using those poles for anything else at the time and it looked untidy. I was envisaging the second poles driving a relay that disconnects the bleed resistor when the system is running.

Using my imagination I've imagined some of that same 24v running thru E-stop going thru limit switch's as well and the RL2 relay having another contact for the 5v signal from BOB.!!

Running the limit switches off the E-stop wire sounds like a good idea - they are effectively doing the same thing as an E-stop, after all. However, I've come across a few discussions about the BoB, with several people saying that it would be better not to cut power to it.

Edit: Oh and you'll need an limit override momentry button in there as well to allow reversing off the switch's.

I guess that might help... :cool:

irving2008

20-09-2012, 08:59 AM

An easy way to provide the limit override (non-locking) and start (locking) logic is with a diode:

6929

An alternative would be to use a seperate contact on Relay 2 to enable Relay 1, but I prefer to keep contacts free for use.

6930

Rogue

20-09-2012, 12:02 PM

Just out of interest, can multiple contacts be wired together, ie a 4 pole with contacts rated for 10VDC@10A wired in series to make effectively a single contact of 40VDC@10A? Or would teeny tiny difference in switching speeds cause one set of contacts to close first and take the full brunt before the other contacts caught up, leading to melting, fusing and the opening of portals to alien dimensions?

I ask because one of the plans had a relay for the bleed resistor on the 70V circuit, but I've not seen many (well, any) relays with 70VDC contacts.

Edit: Thinking about it, voltage increases in series - if it's in series then it doesn't matter if they all close at exactly the same time or not.

irving2008

20-09-2012, 01:13 PM

Voltage on relay contacts is generally not important, its the current carrying capacity. If you need more current then you wire the contacts in parallel. The reason why contacts have a lower DC voltage rating is to do with the arcing that happens when the contact opens. Generally the contact gap is very small, maybe <0.5mm or so. With a DC current the arc ionises the air in the gap and causes a plasma to develop. If the voltage is too high this plasma will continue to pass current until the contacts have opened sufficiently that the voltage cant sustain the plasma, which seriously shortens the contact life.

One solution to the arcing is to put a snubber across the relay contacts - a capacitor and resistor in series - the idea being that as the contact opens the capacitor charges up thus drawing current off the contacts and stopping the arc forming until the contacts are far enough apart to prevent the arc being sustained. This alows the contacts to be used at a higher voltage at the rated current.

This isnt a problem with AC contacts as the voltage returns to zero every cycle thus quenching the arc, thus they can be rated at much higher voltages.

Now, in your situation, using it for the bleed resistor isn't a problem. The current passed is tiny (30mA) so the voltage rating of the contacts isn't an issue as very little arc will be generated.

Incidentally, where did you get the 10v from? If you look at graph H55 on the spec (http://www.rapidonline.com/pdf/60-1305e.pdf)for those relays, you will see that the 4PCO version is rated at 30v 6A. At 70v its derated to 0.5A... but thats only to meet the contact lifetime figures at 1800 cycles/hour. If you switch a dozen times a day it'll be a lot!!! On the AC side graph F55/1 shows that at 900VA the contact life is 200,000 cycles at 1800 cycles/hour so its not going to fail in your lifetime (about 45years at 12 cycles/day) lol.

Rogue

20-09-2012, 01:33 PM

While I wanted to keep everything from the same source for convenience, it looks like RS was my next stop.

As far as I can see, this relay should work in any of the proposed situations that involves relays powered from the 24VDC system. As DPDT it is straightforward to wire it as normally open/closed as required.

As far as mounting goes... I could hot glue the relay somewhere and fix wires directly to the contacts. This would mean glueing it upsidedown or on its side. For various reasons I don't really like this idea. There are are DIN sockets available (699-6881) (http://uk.rs-online.com/web/p/relay-sockets/6996881/?origin=PSF_428009%7Cacc) that should be much easier to fix somewhere, and would give me screw terminals to work with. While it adds on another £3 per relay I think that it might save me a bit of worry.

Now I'm up to (in effect) £10 per relay. Hmm.

Let's look back at Irving2008's previous post with drawings. In the second drawing, Relay 1 is handling the AC current and needs to be suitably higher spec. Relay 2, however, is only handling 24V. A 24VDC DPDT relay rated to 220VDC @ 2A (619-3013) (http://uk.rs-online.com/web/p/non-latching-relays/6193013/) is a much more comfortable £1.78 + VAT. Seeing as the transformer isn't even supplying 2A in the first place, presumably that would be sufficient?

Since I started to write this, I note that Irving2008 has posted again. If "..the current passed [for the bleed resistor] is tiny (30mA) so the voltage rating of the contacts isn't an issue as very little arc will be generated", then would the 2A relay above also work in that situation?

irving2008

20-09-2012, 01:57 PM

Yes. The originally spec'd relays (Rapid 60-1310 (http://www.rapidonline.com/Cables-Connectors/Finder-24VDC-Relay-4PCO-60-1310?utm_source=AffWin&utm_medium=Affiliate&awc=1799_1348144738_dcbeb77b26c100d8f25508b9e926ad 25)) were fine for both positions.. I would socket mount them.

The RS699-6869 is overkill with the inbuild LED etc.

The RS619-3013 would be fine for Relay 2, but as a PCB mounted device its hard to use.

If price is critical then consider using Rapid 60-1667 (http://www.rapidonline.com/Electronic-Components/4pco-5a-Power-Relay-24v-Dc-60-1667) in sockets. While these are only rated at 240vAC/5A and 600VA for the switching rate you will use them for they'll be fine and if you're paranoid you could wire the contacts in pairs.

Rogue

20-09-2012, 02:01 PM

Incidentally, where did you get the 10v from? If you look at graph H55 on the spec (http://www.rapidonline.com/pdf/60-1305e.pdf)for those relays, you will see that the 4PCO version is rated at 30v 6A. At 70v its derated to 0.5A... but thats only to meet the contact lifetime figures at 1800 cycles/hour. If you switch a dozen times a day it'll be a lot!!! On the AC side graph F55/1 shows that at 900VA the contact life is 200,000 cycles at 1800 cycles/hour so its not going to fail in your lifetime (about 45years at 12 cycles/day) lol.

The 10V was pulled out of the air for simplicity and example, I wasn't referring to any specific bit of kit.

As for the graphs, the issue for me is translating what I can see into usable information and that is where my lack of knowledge shines through. For example, you turn the supply on and run it for an hour. For that hour you're drawing a fairly constant 5A. My understanding was that, while the contact only switches once, it still has to carry 5A for that hour. My choices were made on that basis. Clearly that basis was wrong which is leading me to make some less-efficient choices.

Rogue

20-09-2012, 02:28 PM

Yes. The originally spec'd relays (Rapid 60-1310 (http://www.rapidonline.com/Cables-Connectors/Finder-24VDC-Relay-4PCO-60-1310?utm_source=AffWin&utm_medium=Affiliate&awc=1799_1348144738_dcbeb77b26c100d8f25508b9e926ad 25)) were fine for both positions.. I would socket mount them.

I'm glad to hear it, as keeping everything to one supplier makes life easier for me. However, I only found reference to it supporting DC coil, not DC contact, which is why I went looking elsewhere.

The RS699-6869 is overkill with the inbuild LED etc.

True, but the lower end ones are out of stock until January. I'm not bothered by LEDs and test buttons as I don't anticipate seeing it once the cover is closed! As the Rapid version is fine then it's a problem solved (or at least avoided) for today.

The RS619-3013 would be fine for Relay 2, but as a PCB mounted device its hard to use.

Fair enough. At this stage I'm keenly aware of wiring convenience.

If price is critical then consider using Rapid 60-1667 (http://www.rapidonline.com/Electronic-Components/4pco-5a-Power-Relay-24v-Dc-60-1667) in sockets. While these are only rated at 240vAC/5A and 600VA for the switching rate you will use them for they'll be fine and if you're paranoid you could wire the contacts in pairs.

Price is critical over all. I don't want to cut corners but I lack the experience to know which corners can be rounded gently and which can be no more than gently tickled with a chamois cloth.

Rogue

15-08-2013, 08:51 PM

Never being one to rush things, I've now got a toroidal transformer sitting on the bench ready to get wired up!

on the Primary side, I want to wire the Grey and Violet wires together, thereby wiring it in series to handle 230Vac from UK mains,
the Blue wire is now 0V and the Brown wire is now 230Vac, then
on the Secondary side, I want to wire the Orange and Black together, then the Yellow and Red together, and
the Orange/Black is now 0V and the Yellow/Red is now 50Vac.

Wiring it in parallel will give me 50Vac @ 12.5A.
After being rectified this will give me 70V @ 12.5A (less voltage drop across the bridge rectifier)

Is anyone happy to cast their eye over the datasheet and confirm that I'm connecting up the right wires before this goes anywhere near the mains? The instructions look straightforward, but I've said that about lots of things that later went horribly wrong :joker:

The original drivers I was looking at from Zapp are now discontinued. The alternative (DM856 (http://www.zappautomation.co.uk/en/digital-stepper-drivers/738-dm856-microstepping-driver.html)) has a max supply voltage of +80, which I understand to be a comfortable margin for this PSU. The closest driver current options, however, are slightly different - 3.8A or 4.3A, whereas my steppers are 4.2A. The old PM752's would provide 4.02A as the closest fit. Is there a significant enough difference in the current provided that would make it worthwhile hunting for different drivers, or is 3.8A fine? These will now be driving an X2, which I'll be using as a stepping stone to my initial DIY design.

The good news is that I should be finished by 2016, possibly late 2015 if I pull my finger out...

JAZZCNC

15-08-2013, 09:24 PM

The original drivers I was looking at from Zapp are now discontinued. The alternative (DM856 (http://www.zappautomation.co.uk/en/digital-stepper-drivers/738-dm856-microstepping-driver.html)) has a max supply voltage of +80, which I understand to be a comfortable margin for this PSU. The closest driver current options, however, are slightly different - 3.8A or 4.3A, whereas my steppers are 4.2A. The old PM752's would provide 4.02A as the closest fit. Is there a significant enough difference in the current provided that would make it worthwhile hunting for different drivers, or is 3.8A fine? These will now be driving an X2, which I'll be using as a stepping stone to my initial DIY design.

Spend little more and Get the AM882 with lead and you can program them to exact amps and much more, they are great drives.

Clive S

15-08-2013, 09:25 PM

Wiring it in parallel will give me 50Vac @ 12.5A.
After being rectified this will give me 70V @ 12.5A (less voltage drop across the bridge rectifier)

..

A small problem with wiring the secondary in parallel is that if the amount of turns on them is not exactly the same the transformer is not as efficient.
I got around this by rectifying the two windings separately then combining them at the DC side.
Just a thought. ..Clive

Rogue

15-08-2013, 10:13 PM

Spend little more and Get the AM882 with lead and you can program them to exact amps and much more, they are great drives.

Intriguing. I've seen a bit posted about the AM882 drivers but a lot of it didn't seem positive. On some further reading, most of the issues seem to be down to user error and misconfiguration. The price difference at this stage is minimal as well.

Is it better to match up the driver output with the motor rating exactly, or leave a bit of a margin?

A small problem with wiring the secondary in parallel is that if the amount of turns on them is not exactly the same the transformer is not as efficient.
I got around this by rectifying the two windings separately then combining them at the DC side.
Just a thought. ..Clive

Also intriguing, though how would you establish that? Would it be as straightforward as checking the voltage from each output separately? I suspect I will take the easy way out and just use the one rectifier, but I'm intrigued by the idea.

Jonathan

15-08-2013, 10:17 PM

A small problem with wiring the secondary in parallel is that if the amount of turns on them is not exactly the same the transformer is not as efficient.
I got around this by rectifying the two windings separately then combining them at the DC side.
Just a thought. ..Clive

If they're mismatched that configuration will still draw less current from the winding with a lower emf.

The following quick simulations I just did demonstrates this:

9507

9508

The first graph shows the current drawn from each winding for the aforementioned mismatched transformer with a light load and the second graph shows the same supply with a greater load. This shows that the effect if greatest when the load current is low, so although the currents are mismatched they will both be within the rating for a light load. Similarly, the mismatch is smaller when the load is greater, so if the rating of one winding is exceeded it is unlikely to be by enough to cause a problem. Also, there are a lot of other factors besides load current involved here, so the above simulation is by no means comprehensive.

In short it's extremely unlikely for the windings to be mismatched enough for either method to cause problems, so although I'd advise buying a transformer rated for half the voltage so you can put the secondarys in series, if you find one for the right price which has to be wired in parallel I wouldn't worry.

Clive S

15-08-2013, 10:20 PM

Also intriguing, though how would you establish that? Would it be as straightforward as checking the voltage from each output separately? I suspect I will take the easy way out and just use the one rectifier, but I'm intrigued by the idea.

I have just joined the two outputs from the caps together with no problems at all. ..Clive

Jonathan

15-08-2013, 10:22 PM

Is it better to match up the driver output with the motor rating exactly, or leave a bit of a margin?

In theory you should be safe to run any driver up to its rating, since that's the whole point of a rating, if you can trust it. In reality however it's always best to overrate the drivers if you can, since then they will run at a lower temperature which prolongs the life of the components.

Also intriguing, though how would you establish that? Would it be as straightforward as checking the voltage from each output separately?

Yes - the difficult bit is determining how much of a mismatch would pose a problem.

JAZZCNC

15-08-2013, 10:31 PM

In short it's extremely unlikely for the windings to be mismatched enough for either method to cause problems, so although I'd advise buying a transformer rated for half the voltage so you can put the secondarys in series, if you find one for the right price which has to be wired in parallel I wouldn't worry.

Agree here and I've built both ways and never seen any difference to machine which is what matters at end of day. I buy what's cheapest or available and don't care which.

Intriguing. I've seen a bit posted about the AM882 drivers but a lot of it didn't seem positive. On some further reading, most of the issues seem to be down to user error and misconfiguration. The price difference at this stage is minimal as well.

Is it better to match up the driver output with the motor rating exactly, or leave a bit of a margin?

Yes there have been but that's mostly been because they have little experience or never used digital drives before but trust me they are fantastic drives and give super smooth performance with great resonance handling. The Fact you can configure them thru software helps if you have problems with resonance or just want to get the best performance you can.

Yes it's always better to match current to motor ratings but not go higher. The only margin you want to leave is on the voltage but it's also always better if the drives are not max'd out on current IE max current setting = motor current so the drives not working flat out.!

Rogue

15-08-2013, 10:31 PM

In theory you should be safe to run any driver up to its rating, since that's the whole point of a rating, if you can trust it. In reality however it's always best to overrate the drivers if you can, since then they will run at a lower temperature which prolongs the life of the components.

Hi Jonathan, thanks for chipping in again. I'm still aiming to finish this before you finish at Uni so I'll need you to work towards a PhD, should give me enough time.

I was thinking about this from the point of view of the stepper motors (4.2A) rather than the drivers. Either of the two drivers discussed are rated higher than the stepper (up to 5.2A for the DM856, up to 8.2A for the AM882) so there is plenty of margin. My question was whether it's better to run the steppers at their max or give them a bit of a margin. If your answer still applies (which logic suggests it does) then it seems reasonable to set the drivers for a little under 4.2A.

Rogue

15-08-2013, 10:36 PM

Yes it's always better to match current to motor ratings but not go higher. The only margin you want to leave is on the voltage but it's also always better if the drives are not max'd out on current IE max current setting = motor current so the drives not working flat out.!

That answers that question, I should have waited another minute or two before replying! There's approximately 10V difference between the transformer output and the driver max input which should hopefully be fine.

JAZZCNC

15-08-2013, 10:49 PM

That answers that question, I should have waited another minute or two before replying! There's approximately 10V difference between the transformer output and the driver max input which should hopefully be fine.

Yes 10V will be fine and the fact your using toroidal transformer helps further still because the caps will absorb back EMF better.

Jonathan

15-08-2013, 11:01 PM

Hi Jonathan, thanks for chipping in again. I'm still aiming to finish this before you finish at Uni so I'll need you to work towards a PhD, should give me enough time.

That's the plan, however at work they seem to want me to do the PhD whilst working them, in which case you've only got a year left!

My question was whether it's better to run the steppers at their max or give them a bit of a margin. If your answer still applies (which logic suggests it does) then it seems reasonable to set the drivers for a little under 4.2A.

Yes my answer does still apply. As you increase the current in the motor windings, the copper losses increase by the square of the current, so the motor will get hotter. The degradation of the insulation on the motor windings is dependent on the temperature, as with most chemical reactions, so according to the Arrhenius equation we can say (very approximately) that if the motors run 10° hotter, their lifetime will be halved (assuming they fail due to insulation breakdown). This sounds bad, but we have to put it in perspective - if the motors last 20 years then halving this to 10 years probably isn't a big deal.
I'd set the driver current as near as you can to the motor rating without exceeding it. You might be able to go lower, but I think it's better to ensure the machine will not stall by having a good safety margin, rather than worrying about the lifetime of the motors.

Going back to me previous point about the power supply, I just ran the simulation again but this time with both configurations. The first graph is with the windings connected in parallel on the DC side and the second graph has them connected in parallel before the rectifier:

9509

I realise that there's not much point discussing this further, but it's interesting to note that the waveforms do look slightly better for the DC coupled version. There's still not much in it.

m.marino

17-08-2013, 08:31 PM

AM882's are an excellent bit of kit and really are easy to use once you get around the chinese strange way of setting up the software for programing the drives. I use them and they really do make things run very very smoothly. -Michael

Rogue

17-08-2013, 09:46 PM

That's the plan, however at work they seem to want me to do the PhD whilst working them, in which case you've only got a year left!

Glad to hear it, it's not that common these days to get your foot in the door of a good job straight out of Uni. Make the most of it and don't forget us whilst you're busy making money!

AM882's are an excellent bit of kit and really are easy to use once you get around the chinese strange way of setting up the software for programing the drives. I use them and they really do make things run very very smoothly. -Michael

From reading around a bit more I think I'm sold on the AM882 drivers. Currently got 3 of them (+ a lead) sitting in my basket at Zapp. Wisdom dictates that I don't pull the trigger until the end of the month, but my trigger finger is getting fairly itchy at this stage :whistle:

JAZZCNC

17-08-2013, 11:09 PM

From reading around a bit more I think I'm sold on the AM882 drivers. Currently got 3 of them (+ a lead) sitting in my basket at Zapp. Wisdom dictates that I don't pull the trigger until the end of the month, but my trigger finger is getting fairly itchy at this stage :whistle:

Do it because sure has Egg's is egg's they'll be out of stock when you want them.! . . PUSH THE BUTTON!!

Rogue

18-08-2013, 12:24 AM

Do it because sure has Egg's is egg's they'll be out of stock when you want them.! . . PUSH THE BUTTON!!

Don't tempt me! I really, really shouldn't. If the drivers are out of stock then I can always move the ballscrew purchase forward by a month.

Swarfing

18-08-2013, 12:28 AM

House prices are rising, everything else will follow?

Rogue

18-08-2013, 12:33 AM

House prices are rising, everything else will follow?

That is very possible but if an unexpected expense crops up and I've already cut into my "emergency petrol money", then I can't travel. If I can't travel, I can't work and can't invoice clients. That makes my bank manager sad and my wife furious :whistle:

JAZZCNC

18-08-2013, 12:40 AM

That makes my bank manager sad and my wife furious :whistle:

That's easy sorted.?? . . . .change both.!!

Now on another serious note.? if you haven't bought the screws yet then you don't want to be buying the electrics. Only buy the electrics when you need to fit them has just having them to spin on the bench is wasting the warranty and you won't learn anything from it.!!

Swarfing

18-08-2013, 12:41 AM

Rouge the only thing i can say to that is what is more important you? On another site i saws omething that i liked, just came out on the market. I decided then and there to buy it. 6 hours later they doubled the price? and that was from China, go figure?

I know what you mean about work though it has to come first so it sounds like you have answered your own question and it will have to stay on hold. Hope you get there soon, bank managers are a PITA to please.

Rogue

18-08-2013, 02:07 AM

Now on another serious note.? if you haven't bought the screws yet then you don't want to be buying the electrics. Only buy the electrics when you need to fit them has just having them to spin on the bench is wasting the warranty and you won't learn anything from it.!!

Wise words indeed, though (i) the steppers are already outside of a year old and (ii) in any event I've got 6 years to bring a claim for breach of SoGA; warranties do not do what most people think they do. Entirely academic as I'd not pursue that route for a hundred quid's worth of motors, and probably not even for three hundred quid's worth of drivers. Point taken, though, hassle is worth avoiding where possible.

In fact, taking that on board, I might shift my plans around a bit. I'll still finish up the PSU first because it's nearly finished anyway!

Robin Hewitt

18-08-2013, 09:18 AM

In fact, taking that on board, I might shift my plans around a bit. I'll still finish up the PSU first because it's nearly finished anyway!

If you have already bought the toroidal I say go for it, something else could crock the project and fixing the PSU would only waste money.
.
The problem with transformers is that when you rectify them to DC the voltage drops when you draw power. You can stuff capacitors in until the cows come home but if you don't rectify all 3 phases, the volts will sag.
.
Similarly the volts will soar if you get a mains surge so you really need to protect your drivers if they don't protect themselves. Fortunately protection is cheap and easy if you use a crowbar circuit and wire a diac to blow a fuse on the DC side if you go over volts. A crowbar circuit is cheap and simple. Leave yourself a note by the fuse because it is very easy to forget it is there even after it just saved you hundreds of pounds :rugby:

Rogue

11-01-2014, 05:00 PM

Quick question in relation to a soft start circuit on the 70V system.

In all the various circuits I've seen, there is some form of resistance inline before the transformer which, after a very short delay, is then bypassed. Many circuits also then isolate the resistors, though there seems to be various thoughts on whether this is really necessary.

My question relates to using a DPDT relay for this: when the relay activates, the first pole (NO) is closed, creating the bypass. The second pole (NC) which leads to the resistors is opened, taking them out of the circuit. In this arrangement, there is presumably the possibility that the second pole could open before the first pole has closed, so there is no power flowing at all for a very brief time. If this is correct, is this likely to have a significant impact given the timescales involved? Is any switching delay likely to be long enough to negate the effect of the soft start circuit?

If not practical then fair enough, but it would be neater to use one relay to do both functions!

Jonathan

11-01-2014, 05:19 PM

If the rating of your transformer is less than around 600VA, I wouldn't bother with any soft start circuit.
Thermistor is an easy option, as it's resistance decreases with temperature so no real need to take it out of the circuit.
You can use a relay to short circuit the fixed resistor. Essentially you want to switch the power on, wait about a second, then switch the relay on. You've reminded me of a circuit I came up with when I was about 10 - put a largeish capacitor in parallel with the relay coil and charge it via a resistor, then wire the relay output to break the capacitor charging current and you end up with a simple oscillator! In your case you would just use the capacitor and resistor to make a 1 second delay, so calculate it such that it takes 1 second for the capacitor voltage to reach the threshold voltage of the relay coil. Remember to include the resistance of the relay coil in parrallel with the capacitor in your calculation. You can power the relay coil from the 70V output via a potential divider. It's a bit of a crude way to do it, but easy to understand. Bonus points for using a solid state relay.

Rogue

11-01-2014, 06:43 PM

If the rating of your transformer is less than around 600VA, I wouldn't bother with any soft start circuit.
Thermistor is an easy option, as it's resistance decreases with temperature so no real need to take it out of the circuit.

It's 625VA. It might not be very useful but it's more useful than detrimental, I suppose.

I understood the problems with thermistors to be the time needed to cool down again if they are not taken out of the circuit - if the PSU is shut off but restarted within quite a short time, the thermistor doesn't offer enough resistance to work as an effective soft start.

...In your case you would just use the capacitor and resistor to make a 1 second delay, so calculate it such that it takes 1 second for the capacitor voltage to reach the threshold voltage of the relay coil. ... You can power the relay coil from the 70V output via a potential divider.

There's 24V available for such electrickery, though I suppose the more self-contained the system the better.

m_c

11-01-2014, 06:44 PM

Alternatively, use a zero crossing point SSR to power on the transformer, which greatly reduces any switch on surge from the transformer itself. And if you really find a series resistor necessary, just use a resistor bypass relay powered from the DC side. By the time the DC is high enough to power the relay, any surge from bypassing the resistor should be pretty minimal.

EddyCurrent

11-01-2014, 06:51 PM

I'm reading everywhere that zero crossing is the worst thing with regard to inrush.

Jonathan

11-01-2014, 07:06 PM

It's 625VA. It might not be very useful but it's more useful than detrimental, I suppose.

I understood the problems with thermistors to be the time needed to cool down again if they are not taken out of the circuit - if the PSU is shut off but restarted within quite a short time, the thermistor doesn't offer enough resistance to work as an effective soft start.

You would be fine with 625VA and no soft start circuit, so if you restart the PSU with the thermistor still hot the current is still going to be less than without anything, so either way it's not a problem.

I'm reading everywhere that zero crossing is the worst thing with regard to inrush.

Reference?

if you really find a series resistor necessary, just use a resistor bypass relay powered from the DC side. By the time the DC is high enough to power the relay, any surge from bypassing the resistor should be pretty minimal.

I used that strategy recently to charge a 0.06F capacitor bank from the mains via voltage doubler ... nice and simple. Pity what it was powering blew up.

m_c

11-01-2014, 07:09 PM

I'm reading everywhere that zero crossing is the worst thing with regard to inrush.

Not when starting something like a transformer. A mains powered transformer essentially has a zero current point 100 times a second, so if you can switch on at exactly that point, you won't see any surge on the input line, unlike if you were to switch it on when mains volatage is at it's peak you'll see a major surge as full peak voltage is essentially shorted out until the magnetic field builds up.

The issue with surges comes when you add things like capacitors, as these need time to charge. For smaller transformers, the transformer itself will usually limit current enough to avoid any major surge issues, however with larger transformers they can allow that much current to pass that you need to limit it until things get charged up.

m_c

11-01-2014, 07:09 PM

I used that strategy recently to charge a 0.06F capacitor bank from the mains via voltage doubler ... nice and simple. Pity what it was powering blew up.

There is a lot to be said for the KISS approach.
No point complicating things anymore than what you really have to!

EddyCurrent

11-01-2014, 07:20 PM

Reference?

I've known about zero switching for many years and was surprised myself.

You would be fine with 625VA and no soft start circuit, so if you restart the PSU with the thermistor still hot the current is still going to be less than without anything, so either way it's not a problem.

What is the sort of level where soft start becomes important? I kept hitting the figure 500VA when reading, which is why I was looking at including it. Are there other factors that might impact on the choice, ie running it off a standard domestic wall socket, etc?

Jonathan

11-01-2014, 11:00 PM

Not when starting something like a transformer. A mains powered transformer essentially has a zero current point 100 times a second, so if you can switch on at exactly that point, you won't see any surge on the input line, unlike if you were to switch it on when mains volatage is at it's peak you'll see a major surge as full peak voltage is essentially shorted out until the magnetic field builds up.

That is true for a pure resistance, however the issue is the transformer in this case is better modeled as an inductor. I followed Eddy's links but unfortunately they only give 'hand waving' explanations, so instead I've just calculated it and it transpires that if you model the transformer as such, using Faraday's law to find the voltage across the inductor as a function of magnetising flux, then solve this differential equation to find the magnetising flux as a function of applied voltage and switching angle, the result is sinosoidal with a DC-offset. This DC offset depends on the initial flux (i.e. residual flux) and the cosine of the switching angle, so clearly if the angle is pi/2 (i.e. a voltage peak), the cosine term is zero and you get the lowest inrush current. To make matters worse, the relationship between flux and current will be non linear since in normal operation the core operates adjacent to saturation, so when switched on the core is operating well into the saturation region. This means that although the flux implied by Faraday's law is only up to twice the rated value, the current is many times higher. Unfortunately things change a bit when you have capacitors connected to the output via a rectifier, as they essentially present a short circuit to the secondary.

What is the sort of level where soft start becomes important? I kept hitting the figure 500VA when reading, which is why I was looking at including it. Are there other factors that might impact on the choice, ie running it off a standard domestic wall socket, etc?

To be honest my 600VA figure was just a rough estimate based on experience. Yes, you could consider the current capacity of the mains circuit you are connecting it to - e.g. if it's got a 30A RCD with no significant load connected then you're much less likely to have a problem, as the surge current trip is much greater than 30A...

JAZZCNC

11-01-2014, 11:58 PM

To be honest my 600VA figure was just a rough estimate based on experience. Yes, you could consider the current capacity of the mains circuit you are connecting it to - e.g. if it's got a 30A RCD with no significant load connected then you're much less likely to have a problem, as the surge current trip is much greater than 30A...

I'll back up this based off experience because I often use 500 & 600Va transformers without any trouble with inrush trips. If there ever is I just use C or D rated MCB and I'm sorted.

Good article, but they could have expressed the final equation more clearly to demonstrate their point. Also, it's a bit pointless to solve the differential equation using Laplace transforms when you can trivially separate the variables. This is how I did it:

From Faraday's law:
\frac{d\phi}{dt}=\frac{V_p}{N}\sin({\omega}t+\thet a})
Seperate variables:
\int\frac{V_p}{N}\sin({\omega}t+\theta})=\int{d\th eta}
Use sin(A+B) identity to make it easy to integrate:
\frac{V_p}{N}\left[\sin({\omega}t)\cos(\theta)+cos({\omega}t)\sin(\th eta)\right]dt=\phi(t)
Integrate:
\frac{V_p}{\omega{N}}\left[-\cos({\omega}t)\cos(\theta)+sin({\omega}t)\sin(\th eta)\right]+k=\phi(t)
Use cos(A+B) identity to simplify:
\phi(t)=-\frac{V_p}{\omega{N}}\cos({\omega}t+\theta})+k (1)
When the transformer is switched on, we have the initial condition relating to the residual flux:
at t=0, \phi(t)=0
So substitute this in to (1) to find constant, k:
k=\frac{V_p}{\omega{N}}\cos(\theta)+\phi(0)
So the solution is:
\displaystyle\phi(t)=\frac{V_p}{\omega{N}}\left[\cos(\theta)-\cos({\omega}t+\theta})\right]+\phi(0)

So now it's easier to see that the cos(\theta) term is a constant, which introduces a DC offset that disappears when the switching angle is \theta=\frac{\pi}{2} ... i.e at the peak voltage.

EddyCurrent

12-01-2014, 11:25 PM

I used to like Laplace transforms but that was 40 years ago, I've never needed to use them since :hysterical:

JAZZCNC

12-01-2014, 11:37 PM

I used to like Laplace transforms but that was 40 years ago, I've never needed to use them since :hysterical:

I much prefer the Lapdance performs. . Makes me constant, which introduces a DC offset that disappears when the switching angle is ...at the peak. .:whistle: